JEE Main · 2023 · Shift-IIeasyPB11-050

The correct order of bond enthalpy (kJ\,mol-1) is:

p-Block Elements (Class 11) · Class 11 · JEE Main Previous Year Question

Question

The correct order of bond enthalpy (kJ mol−1\mathrm{kJ\,mol^{-1}}) is:

Options
  1. a

    Si−Si>C−C>Sn−Sn>Ge−Ge\mathrm{Si-Si > C-C > Sn-Sn > Ge-Ge}

  2. b

    Si−Si>C−C>Ge−Ge>Sn−Sn\mathrm{Si-Si > C-C > Ge-Ge > Sn-Sn}

  3. c

    C−C>Si−Si>Sn−Sn>Ge−Ge\mathrm{C-C > Si-Si > Sn-Sn > Ge-Ge}

  4. d

    C−C>Si−Si>Ge−Ge>Sn−Sn\mathrm{C-C > Si-Si > Ge-Ge > Sn-Sn}

    ✓
Correct Answerd

C−C>Si−Si>Ge−Ge>Sn−Sn\mathrm{C-C > Si-Si > Ge-Ge > Sn-Sn}

Detailed Solution

Strategy: Rank Group 14 M–M bond enthalpies and determine the correct order.

Step 1: Bond Enthalpy Values (kJ/mol) | Bond | Enthalpy (kJ/mol) | |-------|-----------------| | C–C | ~346 | | Si–Si | ~226 | | Ge–Ge | ~188 | | Sn–Sn | ~151 |

As atomic size increases down the group, orbital overlap decreases, bond length increases, and bond enthalpy decreases.

Step 2: Correct Order \ceC−C>Si−Si>Ge−Ge>Sn−Sn\ce{C-C > Si-Si > Ge-Ge > Sn-Sn}

Answer: (D) \ceC−C>Si−Si>Ge−Ge>Sn−Sn\boxed{\text{Answer: (D) }\ce{C-C > Si-Si > Ge-Ge > Sn-Sn}}

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